Mahesh

Students can practice 10th Class Maths Solutions Telangana Chapter 5 Quadratic Equations Ex 5.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Exercise 5.1

Question 1.
Check whether the following are quadratic equations or not.
i) (x + 1)² = 2(x – 3)
Solution:
Given: (x + 1)² = 2(x – 3)
⇒ x² + 2x + 1 = 2(x – 3)
⇒ x² + 2x + 1 – 2x + 6 = 0
⇒ x² + 7 = 0 is a Q.E.

ii) x² – 2x = -2(3 – x)
Solution:
Given : x² – 2x = -2(3 – x)
⇒ x² – 2x = -6 + 2x
⇒ x² – 4x + 6 = 0 is a Q.E.

iii) (x – 2) (x + 1) = (x – 1) (x + 3)
Solution:
Given : (x – 2) (x + 1) = (x – 1) (x + 3)
⇒ x(x + 1) -2(x + 1)
= x(x + 3) – 1(x + 3)
Note : Compare the coefficients of x² on both sides. If they are equal it is not a Q.E.
⇒ x² + x – 2x – 2 = x² + 3x – x – 3
⇒ x² – x – 2 = x² + 2x – 3
⇒ 3x – 1 = 0 is not a Q.E.

iv) (x – 3) (2x + 1) = x(x + 5)
Solution:
Given : (x – 3) (2x + 1) = x(x + 5)
⇒ x(2x + 1) – 3(2x + 1) = x. x + 5.x
⇒ 2x² + x – 6x – 3 = x² + 5x
⇒ x² – 10x – 3 = 0 is a Q.E.
(or)
Comparing the coefficients of x² on both sides, x . 2x and x. x ⇒ 2x² and x² 2x² ≠ x²
Hence it’s a Q.E.

v) (2x – 1)(x – 3) = (x + 5)(x – 1)
Solution:
Given : (2x – 1)(x – 3) = (x + 5)(x – 1)
⇒ 2x(x – 3) – 1(x – 3) = x(x – 1)+ 5(x – 1)
⇒ 2x² – 6x – x + 3 = x² – x + 5x – 5
⇒ 2x² – 7x + 3 – x² – 4x + 5 = 0
⇒ x² – 11x + 8 = 0
Hence it’s a Q.E.
(or)
Co.eff. of x² on L.H.S. = 2 × 1 = 2
Co.eff. of x² on R.H.S = 1 × 1 = 1
LHS ≠ RHS
Hence it is a Q.E.

vi) x² + 3x + 1 = (x – 2)²
Solution:
Given : x² + 3x + 1 = (x – 2)²
⇒ x² + 3x + 1 = (x – 2)²
⇒ x² + 3x + 1 = x² – 4x + 4
⇒ 7x – 3 = 0 is not a Q.E.

vii) (x + 2)³ = 2x (x² – 1)
Solution:
Given : (x + 2)³ = 2x(x² – 1)
⇒ x³ + 6x² + 12x + 8 = 2x³ – 2x
[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
⇒ x³ + 6x² + 14x + 8 = 0 is not a Q.E. [∵ degree = 3]

viii) x³ – 4x² – x + 1 = (x – 2)³
Solution:
Given : x³ – 4x² – x + 1 = (x – 2)³
⇒ x³ – 4x² – x + 1 = (x – 2)³
= x³ – 6x² + 12x – 8
⇒ 6x² – 12x + 8 – 4x² – x + 1 = 0
⇒ 2x² – 13x + 9 = 0 is a Q.E.

Question 2.
Represent the following situations in the form-of quadratic equations :
i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Solution:
Let the breadth of the rectangular plot be x m.
Then its length (by problem) = 2x + 1.
Area = l . b = (2x + 1). x = 2x² + x but area = 528 m² (∵ given)
∴ 2x² + x = 528
⇒ 2x² + x – 528 = 0 where x is the breadth of the rectangle.

ii) The product of the consecutive positive integers is 306. We need to find the integers.
Solution:
Let the consecutive integers be x and x + 1
Their product = x(x + 1) = x² + x
By problem x² + x = 306
⇒ x² + x – 306 = 0
Where x is the smaller integer.

iii) Rohan’s mother is 26 years older than him. The product of their ages after 3 years will be 360 years. We need to find Rohan’s present age.
Solution:
Let the present age of Rohan be x years. Then age of Rohan’s mother = x + 26
After 3 years :
Age of Rohan would be = x + 3
Rohan’s mother’s age would be = (x + 26) + 3 = x + 29
By problem (x + 3) (x + 29) = 360
⇒ x(x + 29) + 3(x + 29) = 360
⇒ x² + 29x + 3x + 87 = 360
⇒ x² + 32x + 87 – 360 = 0
⇒ x² + 32x – 273 = 0
Where x is Rohan’s present age.

iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/hr less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution:
Let the speed of the train be x kmph. Then time taken to travel a distance

It the speed is 8km/hr less, then time needed to cover the same distance
would be \(\frac{480}{x-8}\)
By problem = \(\frac{480}{x-8}\) – \(\frac{480}{x}\) = 3

⇒ x² – 8x = 1280
⇒ x² – 8x – 1280 = 0
Where x is the speed of the train.

Students can practice 10th Class Maths Solutions Telangana Chapter 4 Pair of Linear Equations in Two Variables Ex 4.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Exercise 4.3

Question 1.
Solve each of the following pairs of equations by reducing them to a pair of linear equations. (T.S. Mar. 15) (A.P. Mar. 15)

i) \(\frac{5}{x-1}\) + \(\frac{1}{y-2}\) = 2
\(\frac{6}{x-1}\) – \(\frac{3}{y-2}\) = 1
Solution:
Given, \(\frac{5}{x-1}\) + \(\frac{1}{y-2}\) = 2 and
\(\frac{6}{x-1}\) – \(\frac{3}{y-2}\) = 1
Put \(\frac{1}{x-1}\) = a and \(\frac{1}{y-2}\) = b,
then the given equations reduce to
5a + b = 2 —– (1)
6a – 3b = 1 —– (2)

Substituting b = \(\frac{1}{3}\) in equation(1), we get
5a + \(\frac{1}{3}\) = 2 ⇒ 5a = 2 – \(\frac{1}{3}\)
⇒ 5a = \(\frac{3 \times 2-1}{3}\)
⇒ 5a = \(\frac{6-1}{3}\)
∴ a = \(\frac{5}{3}\) × \(\frac{1}{5}\) = \(\frac{1}{3}\)
But a = \(\frac{1}{x-1}\) ⇒ \(\frac{1}{3}\) = \(\frac{1}{x-1}\)
⇒ (x – 1) . 1 = 3 x 1
x – 1 = 3 ⇒ x = 3 + 1 = 4
b = \(\frac{1}{y-2}\) ⇒ \(\frac{1}{3}\) = \(\frac{1}{y-2}\)
⇒ y – 2 = 3 ⇒ y = 3 + 2 = 5
∴ Solution (x, y) = (4, 5)

ii) \(\frac{x+y}{x y}\) = 2, \(\frac{x-y}{x y}\) = 6
Solution:

Then the given equations reduces to

Substituting b = 4 in equation (1), we get
a + 4 = 2 ⇒ a = 2 – 4 = -2
But a = \(\frac{1}{x}\) = -2 ⇒ x = \(\frac{-1}{2}\)
b = \(\frac{1}{y}\) = 4 ⇒ y = \(\frac{1}{4}\)
∴ The solution (x, y) = \(\left[\frac{-1}{2}, \frac{1}{4}\right]\)

iii) \(\frac{2}{\sqrt{x}}\) + \(\frac{3}{\sqrt{y}}\) = 2; \(\frac{4}{\sqrt{x}}\) – \(\frac{9}{\sqrt{y}}\) = -1
Solution:
Given, \(\frac{2}{\sqrt{x}}\) + \(\frac{3}{\sqrt{y}}\) = 2 and \(\frac{4}{\sqrt{x}}\) – \(\frac{9}{\sqrt{y}}\) = -1
Take \(\frac{1}{\sqrt{x}}\) = a and \(\frac{1}{\sqrt{y}}\) = a
Then the given equations reduce to
2a + 3b = 2 —- (1)
4a – 9b = – 1 —– (2)

Substituting, b = \(\frac{1}{3}\) in equation (1), we get

iv) 6x + 3y = 6xy
2x + 4y = 5xy
Solution:
Given, 6x + 3y = 6xy

Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b
3a + 6b = 6 —– (1)
4a + 2b = 5 —- (2)

v) \(\frac{5}{x+y}\) – \(\frac{2}{x-y}\) = -1, \(\frac{15}{x+y}\) + \(\frac{7}{x-y}\) = 10 where x ≠ 0, y ≠ 0.
Solution:
Given, \(\frac{5}{x+y}\) – \(\frac{2}{x-y}\) = -1 and \(\frac{15}{x+y}\) + \(\frac{7}{x-y}\) = 10
Take \(\frac{1}{x+y}\) = a and \(\frac{1}{x-y}\) = b, then the given equations reduce to
5a – 2b = -1 —— (1)
15a + 7b = 10 —- (2)

Substituting b = 1 in equation (1), we get
5a – 2(1) = -1 ⇒ 5a = -1 + 2
5a = 1 ⇒ a = \(\frac{1}{5}\)
But a = \(\frac{1}{x+y}\) = \(\frac{1}{5}\) ⇒ x + y = 5
b = \(\frac{1}{x-y}\) = 1 ⇒ x – y = 1
Solving the above equations

Substituting x = 3 in x + y = 5, we get
3 + y = 5 ⇒ y = 5 – 3 = 2
∴ The solution (x, y) = (3, 2)

vi) \(\frac{2}{x}\) + \(\frac{3}{y}\) = 13; \(\frac{5}{x}\) – \(\frac{4}{y}\) = -2 where x ≠ 0, y ≠ 0. (A.P. Jun. 15)
Solution:
Given, \(\frac{2}{x}\) + \(\frac{3}{y}\) = 13 and \(\frac{5}{x}\) – \(\frac{4}{y}\) = -2
Put \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b
Then the given equations reduce to
2a + 3b = 13 —– (1)
5a – 4b = -2 —– (2)
Solving (1) & (2), we get

Substituting b = 3 in equation (1), we get
2a + 3(3) = 13 ⇒ 2a = 13 – 9
⇒ a = \(\frac{4}{2}\) = 2
But, a = \(\frac{1}{x}\) = 2 ⇒ x = \(\frac{1}{2}\)
b = \(\frac{1}{y}\) = 3 ⇒ y = \(\frac{1}{3}\)
∴ The solution (x, y) = \(\left(\frac{1}{2}, \frac{1}{3}\right)\)

vii) \(\frac{10}{x+y}\) + \(\frac{2}{x-y}\) = 4; \(\frac{15}{x+y}\) – \(\frac{5}{x-y}\) = -2
Solution:
Given, \(\frac{10}{x+y}\) + \(\frac{2}{x-y}\) = 4; \(\frac{15}{x+y}\) – \(\frac{5}{x-y}\) = -2
Taking \(\frac{1}{x+y}\) = a and \(\frac{1}{x-y}\) = b then the given equations reduce to
10a + 2b = 4 —– (1)
15a – 5b = -2 —- (2)

Substituting b = 1 in equation (1), we get 10a + 2(1) = 4
⇒ 10a = 4 – 2 ⇒ a = \(\frac{2}{10}\) = \(\frac{1}{5}\)

Substituting x = 3 in x + y = 5, we get
3 + y = 5 ⇒ y = 5 – 3 = 2
∴ The solution (x, y) = (3, 2)

viii) \(\frac{1}{3 x+y}\) + \(\frac{1}{3 x-y}\) = \(\frac{3}{4}\)
\(\frac{1}{2(3 x+y)}\) – \(\frac{1}{2(3 x-y)}\) = \(\frac{-1}{8}\)
Solution:
\(\frac{1}{3 x+y}\) + \(\frac{1}{3 x-y}\) = \(\frac{3}{4}\) and \(\frac{1}{2(3 x+y)}\) – \(\frac{1}{2(3 x-y)}\) = \(\frac{-1}{8}\)
Take \(\frac{1}{3 x+y}\) = a; \(\frac{1}{3 x-y}\) = b,
then the given equations reduce to

Sustituting a = \(\frac{1}{4}\) in equation (1), we get

Substituting x = 1 in 3x + y = 4
⇒ 3(1) + y = 4 ⇒ y = 4 – 3 = 1
∴ The solution (x, y) = (1, 1)

Question 2.
Formulate the following problems as a pair of equations and then find their solutions.
i) A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and speed of the boat in still water.
Solution:
Let, the speed of the boat in still water = x kmph
and the speed of the stream = y kmph.
then speed in downstream = x + y Speed in upstream = x – y

then the above equations reduce to
30a + 44b = 10 —– (1)
40a + 55b = 13 —- (2)

Substituting b = \(\frac{1}{11}\) in equation (1), we get
30a + 44(\(\frac{1}{11}\)) = 10
⇒ 30a = 10 – 4 ⇒ a = \(\frac{6}{30}\) = \(\frac{1}{5}\)
but a = \(\frac{1}{x-y}\) = \(\frac{1}{5}\) ⇒ x – y = 5 —- (3)
b = \(\frac{1}{x+y}\) = \(\frac{1}{11}\) ⇒ x + y = 11 —- (4)
Adding (3) and (4)

Substituting x = 8 in x – y = 5, we get
8 – y = 5 ⇒ y = 8 – 5 = 3
∴ The solution (x, y) = (8, 3)
Speed of the boat in still water = 8 kmph
Speed of the stream = 3kmph

ii) Rahim travels 600 km to his home partly by train and partly by car. He takes 8 hours If he travels 120 km by train and rest by car. He takes 20 minutes more If he travels 200 km by train and rest by car. Find the speed of the train and the car.
Solution:
Let, the speed of the train be x kmph, and the speed of the car be y kmph
By problem,

then the equations reduce to
15a + 60b = 1 —- (1)
8a + 16b = \(\frac{1}{3}\) ⇒ 24a + 48b = 1 —- (2)

Substituting a = \(\frac{1}{36}\) in Equation (1)
we get,

∴ Time taken by 1 woman 18 days
Time taken by 1 man = 36 days

Students can practice 10th Class Maths Solutions Telangana Chapter 5 Quadratic Equations Ex 5.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Exercise 5.4

Question 1.
Find the nature of the roots of the following quadratic equations. If real roots exist, find them. (A.P June is)

i) 2x² – 3x + 5 = 0
Solution:
Given : 2x² – 3x + 5 = 0
a = 2; b = -3; c = 5
Discriminant = b² – 4ac
b² – 4ac = (-3)² – 4(2)(5)
= 9 – 40
= -31 < 0
Roots are imaginary,

ii) 3x² – 4 \(\sqrt{3}\)x + 4 = 0
Solution:
Given : 3x² – 4\(\sqrt{3}\)x + 4 = 0
a = 3; b = -4\(\sqrt{3}\) ; c = 4
b² – 4ac = (-\(\sqrt{3}\))² – 4(3)(4)
= 48 – 48 = 0
∴ Roots are real and equal and they are
\(\frac{-b}{2 a}\), \(\frac{-b}{2 a}\)
= \(\frac{-(-4 \sqrt{3})}{2 \times 3}\) = \(\frac{4 \sqrt{3}}{6}\) = \(\frac{2}{\sqrt{3}}\), \(\frac{2}{\sqrt{3}}\)

iii) 2x² – 6x + 30
Solution:
Given : 2x² – 6x + 3 = 0
a = 2; b = -6; c = 3
b² – 4ac = (-6)² – 4(2)(3)
= 36 – 24 = 12 > 0
The roots are real and distinct.
They are

Question 2.
Find the values of k for each of the following quadratic equations so that they have two equal roots.

i) 2x² + kx + 3 = 0
Solution:
Given : 2x² + kx + 3 = 0 has equal roots
∴ b² – 4ac = 0
Here a = 2; b = k; c = 3
b² – 4ac = (k)² – 4(2) (3) = 0
⇒ k² – 24 = 0
⇒ k² = 24
⇒ k = \(\sqrt{24}\)
= ±2 \(\sqrt{6}\)

ii) kx(x – 2) + 6 = 0
Solution:
Given : kx(x – 2) + 6 = 0
kx² – 2kx + 6 = 0
As this Q.E. has equal roots.
b² – 4ac = 0
Here a = k; b = -2k; c = 6
∴ b² – 4ac = (-2k)² – 4(k) (6) = 0
⇒ 4k² – 24 k = 0
⇒ 4k(k – 6) = 0
⇒ k = 0 (or) k – 6 = 0
⇒ k = 0 (or) 6
But k = 0 is trivial
∴ k = 6

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m² ? If so, find its length and breadth. (A.P. June 15)
Solution:
Let the breadth = x m
Then length = 2x m
Area = length × breadth
= x. (2x) = 2x² m²
By problem 2x² = 800
⇒ x² = 400
and x = \(\sqrt{400}\) = ± 20
∴ Breadth x = 20 m and
Length 2x = 2 × 20 = 40 m

Question 4.
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Is the situation possible ? If so, determine their present ages.
Solution:
Let the age one of the two friends be x years.
Then the age of the other = 20 – x
Then, 4 years ago their ages would be (x – 4) and (20 – x – 4) = 16 – x
∴ Problem (x – 4) (16 – x) = 48
⇒ x(16 – x) – 4(16 – x) = 48
⇒ 16x – x² – 64 + 4x = 48
⇒ x² – 20x + 112 = 0
Here a = 1; b = – 20; c = 112
b² – 4ac = (-20)² – 4(1) (112)
= 400 – 448 = -48 < 0
Thus the roots are not real.
∴ The situation is not possible.

Question 5.
Is it possible to design a rectangular park of perimeter 80m and area 400m² ? If so, find its length and breadth.
Solution:
Given : Perimeter of a rectangle 2(1l + b) = 80
⇒ l + b = \(\frac{80}{2}\) = 40 —– (1)
Area of the rectangle, l × b = 400
It possible, let us suppose that length of the rectangle = x m say
Then its breadth by equation (1) = 40 – x
By problem area = x. (40 – x) = 400
⇒ 40x – x² = 400
⇒ x² – 40x + 400 = 0
Here a = 1; b = – 40; c = 400
b² – 4ac = (-40)² – 4(1)(400)
= 1600 – 1600
= 0
∴ The roots are real and equal.
They are \(\frac{-b}{2 a}\), \(\frac{-b}{2 a}\)
i.e., \(\frac{-(-40)}{2 \times 1}\) = \(\frac{40}{2}\) = 20
∴ The dimensions are 20m, 20m.
(∴ The park is in square shape)

Students can practice 10th Class Maths Solutions Telangana Chapter 5 Quadratic Equations Ex 5.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Exercise 5.3

Question 1.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square. (A.P.Mar. 16, 15)

i) 2x² + x – 4 = 0
Solution:
Given : 2x² + x – 4 = 0
⇒ 2x² + x = 4
⇒ (\(\sqrt{2}\))² + x = 4
⇒ (\(\sqrt{2}\)x)² + 2.\(\sqrt{2}\). x. \(\frac{1}{2 \sqrt{2}}\) = 4
Now LHS is in the form a² + 2ab Where b = \(\frac{1}{2 \sqrt{2}}\)
Adding b² = \(\left(\frac{1}{2 \sqrt{2}}\right)^2\) on both sides we get

ii) 4x² + 4\(\sqrt{3}\)x + 3= 0
Solution:
Given : 4x² + 4\(\sqrt{3}\)x + 3 = 0
⇒ 4x² + 4\(\sqrt{3}\)x = – 3
⇒ (2x)² + 2(2x) \(\sqrt{3}\) = -3
LHS is of the form a² + 2ab where b = \(\sqrt{3}\)
∴ Adding b² = (\(\sqrt{3}\))² = 3 on both sides, we get
⇒ (2x)² + 2(2x)(\(\sqrt{3}\)) + (\(\sqrt{3}\))²
= -3 + (\(\sqrt{3}\))²
⇒ (2x + (\(\sqrt{3}\)))² = -3 + 3 = 0
⇒ (2x + \(\sqrt{3}\))² = 0
⇒ 2x + \(\sqrt{3}\) – 0
⇒ 2x + \(\sqrt{3}\) = 0
⇒ 2x = –\(\sqrt{3}\)
⇒ x = \(\frac{-\sqrt{3}}{2}\)
∴ The roots are \(\frac{-\sqrt{3}}{2}\), \(\frac{-\sqrt{3}}{2}\)

iii) 5x² – 7x – 6 = 0
Solution:
The given Q.E. is 5x² – 7x – 6 = 0
⇒ 5x² – 7x = 6
(\(\sqrt{5}\))² – 2\(\sqrt{5}\). x.\(\left(\frac{7}{2 \sqrt{5}}\right)\) = 6
The LHS is of the form a² – 2ab
where b = \(\frac{7}{2 \sqrt{5}}\)
Now adding b² = \(\left(\frac{7}{2 \sqrt{5}}\right)^2\) on both sides we get

Note : If we take the Q.E. as 5x² – 7x + 6 = 0, then we get the T.B. answer.

iv) x² + 5 = -6x
Solution:
The given Q.E. is x² + 5 = -6x
⇒ x² + 6x = -5
⇒ (x)² + 2. (x). 3 = -5
Now L.H.S. is of the form a² + 2ab where b = 3.
Adding b² = 3² on both sides we get
x² + 2(x)(3) + 3² = -5 + 3²
(x + 3)² = -5 + 9 = 4
∴ x + 3 = \(\sqrt{4}\) = ±2
⇒ x = +2 – 3 or -2 -3
= -1 or -5 are the roots of the given Q.E.

Question 2.
Find the roots of the quadratic equations given in Q.l above by applying the quadratic formula,
i) 2x² + x – 4 = 0
Solution:
Comparing this Q.E.with ax² + bx + c = 0
a = 2, b = 1, c = -4

ii) 4x² + 4\(\sqrt{3}\)x + 3 = 0
Solution:
Given : 4x² + 4 \(\sqrt{3}\) x + 3 = 0
Here a = 4, b = \(\sqrt{3}\); c = 3

iii) 5x² – 7x – 6 = 0
Solution:
Given : 5x² – 7x – 6 = 0
Here a = 5, b = -7 and c = -6

iv) x² + 5 = -6x
Solution:
Given : x² + 5 = -6x
⇒ x² + 6x + 5 = 0
Here a = 1; b = 6; c = 5
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

Question 3.
Find the roots of the following equations :

i) x – \(\frac{1}{x}\) = 3, x ≠ 0
Solution:
Given x – \(\frac{1}{x}\) = 3
⇒ \(\frac{x^2-1}{x}\) = 3
⇒ x² – 1 = 3x
⇒ x² – 3x – 1 = 0
Here a = 1; b = -3; c = 1

ii) \(\frac{1}{x+4}\) – \(\frac{1}{x-7}\) = \(\frac{11}{30}\), x ≠ -4, 7
Solution:

⇒ x² – 3x – 28 = -30
⇒ x² – 3x – 28 + 30 = 0
⇒ x² – 3x + 2 = 0
⇒ x² – 2x – x + 2 = 0
⇒ x(x – 2) – 1(x – 2) = 0
⇒ (x – 2) (x – 1) = 0
⇒ x – 2 = 0 (or) x – 1 = O
⇒ x = 2 or x = 1
∴ x = 2 or 1.

Question 4.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\), Find his present age.
Solution:
Let the present age of Rehman be x years.
3 years age Rehman’s age = x – 3 and its reciprocal is \(\frac{1}{x+5}\)
Rehman’s age 5 years from now = x + 5 and its reciprocal is \(\frac{1}{x+5}\)
The sum of the reciprocals

⇒ x² + 2x – 15 = 3(2x + 2)
⇒ x² + 2x – 15 = 6x + 6
⇒ x² + 2x – 15 – 6x – 6 = 0
⇒ x² – 4x – 21 = 0
⇒ x² – 7x + 3x – 21 0
⇒ x(x – 7) + 3(x – 7) = 0
⇒ (x – 7)(x + 3) = 0
⇒ x = -7 = 0 or x + 3 = 0
⇒ x = 7 or x = -3
But x can’t be negative
∴ x = 7
ie., present age of Rehman = 7 years.

Question 5.
In a class test, the sum of Moulika’s marks in Mathematics and English is 30. If she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in the two subjects.
Solution:
Sum of the marks in Mathematics and English = 30
Let Moulika’s marks in Mathematics be x Then her marks in English = 30 – x
If she got 2 more marks in Mathematics, then her marks would be x + 2.
It she got 3 marks less in English then her marks would be 30 – x – 3 = 27 – x
By problem (x + 2) (27 – x ) = 210
⇒ x(27 – x) + 2(27 – x) = 210
⇒ 27x – x² + 54 – 2x = 210
⇒ -x² + 25x + 54 = 210
⇒ x² – 25x – 54 + 210 = 0
⇒ x² – 25x + 156 = 0
⇒ x² – 12x – 13x + 156 = 0
⇒ x(x – 12) – 13(x – 12) = 0
⇒ x – 12 = 0 or x – 13 = 0
⇒ x = 12 or x = 13
If x = 12, then marks in Mathematics =12
English = 30 – 12 = 18
If x = 13, then marks in Mathematics = 13
English = 30 – 13 = 17

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:
Let the shorter side of the rectangular field = x m.
Then its longer side = x + 30 m

The diagonal of a rectangle is also the hypotenuse of the lower triangle.
Here the diagonal = x + 60
∴ By Pythagoras Theorem
(side)² + (side)² = (hypotenuse)²
⇒ (x + 30)² + x² =(x + 60)²
⇒ x² + 60x + 900 + x² = x² + 120 x + 3600
⇒ x² – 60x – 2700 = 0
⇒ x² – 90x + 30x – 2700 = 0
⇒ x(x – 90) + 30 (x – 90) = 0
⇒ (x – 90) (x + 30) = 0
⇒ x – 90 = 0(or) x + 30 = 0
⇒ x = 90 (or) x = -30
But ‘x’ can’t be negative.
∴ x = 90 m
i.e., the shorter side x = 90 m
Longer side x + 30 = 90 + 30 = 120 m.

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:
Let the large number be x.
8 times larger number = square of the small number = 8x
Square of the large number = x²
By problem, x² – 8x = 180
⇒ x² – 8x – 180 = 0
⇒ x² – 18x + 10x – 180 = 0
⇒ x(x – 18) + 10(x – 18) = 0
⇒ (x + 10) (x – 18) = 0
⇒ x + 10 = 0(or) x – 18 = 0
⇒ x = – 10(or) x = 18
If x = 18, then larger number = 18;
(small number)² = 8 × (18) = 144
∴ Small number = \(\sqrt{144}\) = 12
The numbers are 18, 12
Note : Discard x = – 1o.

Question 8.
A train travels 360 km at a uniform speed. It the speed had been 5 km/br more, It would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
The distnce travelled = 360 km.
Let the speed of the train = x kmph.
Time taken to complete a journey =

By problem \(\frac{360}{x}\) – \(\frac{360}{x+5}\) = 1
⇒ 360\(\left(\frac{1}{x}-\frac{1}{x+5}\right)\) = 1
⇒ 360\(\left(\frac{x+5-x}{x(x+5)}\right)\)
⇒ \(\frac{5}{x^2+5 x}\) = \(\frac{1}{360}\)
⇒ x² + 5x = 1800
⇒ x² + 5x – 1800 = 0
⇒ x² + 45x – 40x – 1800 = 0
⇒ x(x + 45) – 40(x + 45) = 0
⇒ (x + 45) (x – 40) = 0
⇒ x + 45 = 0 or x – 40 = 0
⇒ x = – 45 or x = 40
But x can’t be negative
∴ The speed of the train = 40 kmph.

Question 9.
Two water taps together can fill a tank in 9\(\frac{3}{8}\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:
Let the time taken by tap with smaller diameter alone to fill the tank = x hrs.
Then time taken by the tap with larger diameter = (x – 10) hrs.
By problem both taps worked for
9\(\frac{3}{8}\) = \(\frac{75}{8}\) hours


⇒ 75(x – 5) = 4(x² – 10x)
⇒ 4x² – 40x = 75x – 375
⇒ 4x² – 115x + 375 = 0
⇒ 4x² – 100x – 15x + 375 = 0
⇒ 4x(x – 25) – 15(x – 25) = 0
⇒ x – 25 = 0 (or) 4x – 15 = 0
⇒ x = 25 (or) x = \(\frac{15}{4}\)
x can’t be \(\frac{15}{4}\)
∴ x = 25
i.e., Time taken by the first tap = 25hrs
Time taken by the second tap
= 25 – 10 = 15 hrs.

Question 10.
An express train takes 1 hour less then a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of the two trains.
Solution:
Let the speed of the passenger train = x kmph
Then speed of the express train = x + 11 kmph.
Distance travelled = 132 km

⇒ x² + 11x = 132 × 11
⇒ x² + 11x – 1452 = 0
⇒ x² + 44x – 33x – 1452 = 0
⇒ x(x + 44) – 33(x + 44) = 0
⇒ (x + 44) (x – 33) = 0
⇒ x + 44 = 0 (or) x – 33 = 0
⇒ x = -44 (or) x = 33
But x cant be negative.
∴ Speed of the passenger train
= x = 33 kmph.
Speed of the express train
= x + 11 = 44 kmph.

Question 11.
Sum of the areas of two squares is 468 m². If the difference of the perimeters is 24m, find the sides of the two square.
Solution:
Let the side of first square = x m say
Then perimeter of the first square
= 4x [∵ P = 4. side]
By Problem, perimeter of the second square = 4x + 24 (or) 4x – 24
∴ Side of the second square
⇒ \(\frac{4 x+24}{4}\) = \(\frac{4(x+6)}{4}\) = x + 6 (or)
\(\frac{4 x-24}{4}\) = x – 6
Now sum of the areas of the two squares is given as 468 m²
x² + (x + 6)² = 468
x² + x² + 12x + 36 = 468
2x² + 12x + 36 – 468 = 0
2x² + 12x – 432 = 0
x² + 6x – 216 = 0
x² + 18x – 12x – 216 = 0
x(x + 18) – 12(x + 18) = 0
(x + 18) (x – 12) = 0
⇒ x + 18 = 0 (or) x – 12 = 0
⇒ x = – 18 (or) 12
But x can’t be negative
∴ x = 12
i.e., side of the first square = 12
∴ Perimeter = 4 × 12 = 48
∴ Perimeter of the second square = 48 + 24 = 72
∴ Side of the second square = \(\frac{72}{4}\) = 18 m.
(or)
x² + (x – 6)² = 468
x² + x² – 12x + 36 = 468
2x² – 12x – 432 = 0
x² – 6x – 216 = 0
x²– 18x + 12x – 216 = 0
x(x – 18) + 12(x – 18) = 0
(x – 18) (x + 12) = 0
⇒ x – 18 = 0 (or) x + 12 = 0
⇒ x = 18 (or) -12
But x can’t be negative
∴ x = 18
i.e., side of the first square = 18 m
Perimeter = 4 × 18 = 72
Perimeter of the second square
= 72 – 24 = 48
∴ Side of the second square = \(\frac{48}{4}\) = 12 m
i. e., In any way, the sides of the squares are 12m, 18m.

Question 12.
S = ut – \(\frac{1}{2}\)gt² is a formula which gives the distance S in meters travelled by a ball from the thrower’s hands if it is thrown upwards with an initial velocity of u m/s after a time of t seconds, g is the acceleration due to gravity and is 9.8 m/s²
(i) If a ball is thrown upwards at 14 m/s, how high has it gone after 1 second ?
(ii) How long does it take for the ball to reach a height of 5 meters ?
(iii) Why are there two possible times to reach a height of 5 meters ?
Solution:
S = ut – \(\frac{1}{2}\)gt²
i) u = 14 m/s; t = 1 sec; g = 9.8 m/s²
S = 14 × 1 – \(\frac{1}{2}\) × 9.8 × (1)²
= 14 – \(\frac{1}{2}\) × 9.8 = 14 – 4.9 = 9.1m.

ii) S = 5 m.
5 = 14t – 4.9 t²
4.9 t² – 14t + 5 = 0
\(\frac{49}{10}\)t² – 14t + 5 = 0
49t² – 140t + 50 = 0
a = 49, b = -140, c = 50

iii) When the body goes upwards t value is 2.4 sec.
When the body comes downwards t value is 0.4 sec.
∴ So there are two possible times to reach a height of 5 meters.

Question 13.
If a polygon of ‘n’ sides has \(\frac{1}{2}\)n(n – 3) diagonals. How many sides will a polygon having 65 diagonals ? Is there a polygon with 50 diagonals ?
Solution:
Given : Number of diagonals of a polygon with n-sides = \(\frac{\mathrm{n}(\mathrm{n}-3)}{2}\)
No. of diagonals of a given polygon = 65
i.e., \(\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) = 65
where n is number of sides of the polygon
⇒ n² – 3n = 2 × 65
⇒ n² – 3n – 130 = 0
⇒ n² – 13n + 10n – 130 = 0
⇒ n(n – 13) + 10(n – 13) = 0
⇒ (n – 13) (n + 10) = 0
⇒ (n – 13) = 0 (or) n + 10 = 0
⇒ n = 13 (or) n = -10
But n can’t be negative.
∴ n = 13 (i.e.) number of sides = 13.
Also to check 50 as the number of diagonals of a polygon
∴ \(\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) = 50
⇒ n² – 3n = 100
⇒ n² – 3n – 100 = 0
There is no real value of n for which the above equation is satisfied.
∴ There can’t be a polygon with 50 diagonals.

Students can practice 10th Class Maths Solutions Telangana Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Exercise 4.2

Question 1.
The ratio of incomes of two persons is 9: 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save
₹ 2,000 per month, find their monthly income.
Solution:
Let the income of the first person be ₹ 9x and that of the second person be ₹ 7x.
Further, let the expenditure of the first person and the second person be ₹ 4y and 3y respectively.
Then, the saving of first person = 9x – 4y
By problem, 9x – 4y = 2,000 —— (1)
The saving of second person = 7x – 3y
Solving (1) and (2), we get
We equate the coefficients of ‘y’ in the equations.

∴ x = 2,000
Substitute x = 2,000 in equation (1), we get
9 × 2,000 – 4y = 2,000
18,000 – 4y = 2,000
-4y = 2,000 – 18,000 = -16,000
∴ y = \(\frac{-16000}{-4}\) = 4,000
Therefore, monthly income of the first person = ?9x
= ₹ 9 × 2,000
= ₹ 18,000
Monthly income of the second person = ₹ 7x
= ₹ 7 × 2,000
= ₹ 14,000

Question 2.
The sum of a two digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there ?
Solution:
Let the digit at units place be x.
And the digit at tens place be y.
The number will be yx.
The value of the number = y × 10 + x × 1
= 10y + x

Number obtained by reversing the digits = xy
Then the value of the reversed number = 10 × x + y × 1
= 10x + y
By problem, we have
(10y + x) + (10x + y) = 66
⇒ 11x + 11y = 66
⇒ x + y = 6 —– (1)
It is given that the digits differ by 2.
So, x – y = 2 (or) y – x = 2 —- (2)
Solving (1) and (2), we get

Substitute x = 4 in (1), we get
x + y = 6 ⇒ y = 6 – 4 = 2
Substitute y = 4 in (1),
x + y = 6 ⇒ x = 6 – 4 = 2
∴ The required number is 24.
∴ The required number is 42.
There are two numbers (i.e.,) 24 and 42.

Question 3.
The larger of two supplementary angles exceeds the smaller by 18°. Find the angles.
Solution:
Two angles are said to be supplementary if their sum is 180°.
Let the smaller supplementary angle be x°. and the larger supplementary angle be y°.
We know that sum of these two angles is 180°.
∴ x + y = 180° —- (1)
The larger angle exceeds the smaller by 18°. Then, y = x + 18
⇒ -x + y = 18 —- (2)
Solving (1) & (2), we get

Substitute y = 99 in (1), we get
x + 99 = 180
⇒ x = 180 – 99 = 81
Therefore, the required angles are 81°, 99°.

Question 4.
The taxi charges in Hyderabad are fixed along with the charge for the distance covered. For a distance of 10 km., the charge paid is ₹ 220. For a journey of 15 km, the charge paid is ₹ 310.
i) What are the fixed charges and charge per km ?
ii) How much does a penon have to pay for travelling a distance of 25 km?
Solution:
Let the fixed charge be ₹ x.
And charge per km be ₹y.
For a distance of 10km, the charge paid is ₹ 220.
Then x + 10y = 220 —– (1)
For a distance of 15 km, the charge paid is ₹ 310.
Then, x + 15y = 310 —- (2) ;
Solving (1) & (2), we get

Substitute y = 18 in equation (1), we get
x + (10 × 18) = 220
x + 180 = 220
∴ x = 220 – 180 = 40
Fixed charge = ₹ 40
Charge per km = ₹ 18
∴ Charge for 25 km = ₹ 450

Question 5.
A fraction becomes if \(\frac{4}{5}\) is added to both numerator and denominator. If, however, 5 is subtracted from both numerator and denominator, the fraction becomes \(\frac{1}{2}\). What is the fraction ? (AP Mar. 15)
Solution:
Let the fraction be; \(\frac{x}{y}\)
If 1 is added to both numerator and denominator then the fraction = \(\frac{x+1}{y+1}\)
By problem, \(\frac{x+1}{y+1}\) = \(\frac{4}{5}\)
⇒ 5(x + 1)= 4(y + 1)
⇒ 5x + 5 = 4y +4
⇒ 5x – 4y = 4 – 5 = -1
⇒ 5x – 4y = -1 —- (1)
If 5 is subtracted from both numerator and denominator, then
the fraction = \(\frac{x-5}{y-5}\)
By problem, \(\frac{x-5}{y-5}\) = \(\frac{1}{2}\)
⇒ 2(x – 5) = 1(y – 5)
⇒ 2x – 10 = y – 5
⇒ 2x – y = -5 + 10 = 5
⇒ 2x – y = 5 —- (2)
Solving (1) & (2), we get

Substitute x = 7 in (2),
⇒ 2 × 7 – y = 5
⇒ 14 – y = 5
⇒ -y = 5 – 14 = -9
∴ y = 9
∴ The required fraction is \(\frac{7}{9}\).

Question 6.
Places A and B are 1oo km apart on a highway. One car starts from A and another from B at the same time at different speeds. if the cars travel in the same direction, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Solution:
Let the speed of the car that starts from A be x kmph.
Let the speed of the car that starts from B be y kmph.
Distance between A and B is 1oo km.

∴ Speed × Time = Distance travelled
(x – y) × 5 = 100
5x – 5y = 100
⇒ x – y = 20 —- (1)
If the two cars travel in the opposite direction, their relative speed will be (x + y) kmph. It is also given that the cars will meet in 1 hour.
∴ Speed × Time = Distance travelled
(x + y) × 1 = 100
⇒ x + y = 100 —- (2)
Solving (1) & (2), we get

Substitute x = 60 in equation (2),
60 + y = 100
∴ y = 100 – 60 = 40
∴ The speeds of the cars are 60 kmph and 40 kmph.

Question 7.
Two angles are complementary. The larger angle is 3° less than twice the measure of the smaller angle. Find the
measure of each angle.
Solution:
Two angles are said to be complementary if their sum is 90°.
Let the smaller angle be x° and the larger angle y°.
∴ x + y = 90° —- (1)
Given that the larger angle is 30 less than twice the measure of the smaller angle.
∴ y = 2x – 3
⇒ 2x – y = 3 — (2)
Solving (1) & (2), we get

Substitute x = 31 in (1),
Then, 31 + y = 90
∴ y = 90 – 31 = 59
The two angles are 31° and 59°

Question 8.
An algebra textbook has a total of 1382 pages. It is broken up into two parts. The second part of the book has 64 pages
more than the first part. How many pages are in each part of the book ?
Solution:
Let the number of pages in the first part of algebra textbook be ‘x’ and that of second part be ‘y’. The textbook contains 1382 pages.
∴ x + y = 1382 —- (1)
It is given that the second part of the book has 64 pages more than the first part.
∴ y = x + 64
⇒ x – y = – 64 —- (2)
Solving (1) & (2), we get

Substitute x = 659 in (1), we get
659 + y = 1382
∴ y = 1382 – 659 = 723
First part of the book has 659 pages whereas the second part has 723 pages.

Question 9.
A chemist has two solutions of hydrochloric acid in stock. One is 50% solution and the other is 80% solution. How much of each should be used to obtain 100 ml of a 68% solution ?
Solution:
Let 50% solution to be used be ‘x’ ml.
And 80% solution to be used be ‘y’ ml.
Quantity of solution to be obtained = 100 ml
∴ x + y = 100 —– (1)
50% of ‘x’ ml = \(\frac{x \times 50}{100}\) = \(\frac{1}{2} \mathrm{x}\)
80% of ‘y’ ml = \(\frac{\mathrm{y} \times 80}{100}\) = \(\frac{4}{5} \mathrm{y}\)
By problem, \(\frac{1}{2}\)x + \(\frac{4}{5}\)y = 68
Multiplying each terms by 10, we get
[\(\frac{1}{2}\)x × 10] + [\(\frac{4}{5}\)y × 10] = 68 × 10
5x + 8y = 680 —– (2)
Solving equation (1) & (2), we have

∴ y = \(\frac{180}{3}\) = 60
Substitute y = 60 in (1), we have
x + y = 100
x + 60 = 100
∴ x = 100 – 60 = 40
Therefore, 50% solution to be used = 40 ml.
80% solution to be used 60 ml

Question 10.
Suppose you have ₹ 12,000 to invest. You have to invest some amount at 10% and the rest at 15%. How much should be
invested at each rate to yield 12% on the total amount invested?
Solution:
Total amount to be invested = ₹ 12,000/-
Let the amount invested at 10% be ₹ x.
Let the amount invested at 15% be ₹ y.
∴ x + y = ₹ 12,000 —- (1)
Amount yielded at 10% of
x = \(\frac{x \times 10}{100}\) = \(\frac{x}{10}\)
Amount yielded at 15% of
y = \(\frac{y \times 15}{100}\) = \(\frac{3 \mathrm{y}}{20}\)
yielded at 12% of ₹ 12,000
= \(\frac{12000 \times 12}{100}\) = 1440
∴ \(\frac{x}{10}\) + \(\frac{3 y}{20}\) = \(\frac{1440}{1}\)
∴ 2x + 3y = 28800 —- (2)

Substitute y = 4800 in (1), we get
x + 4800 = 12000
∴ x = 12000 – 4800 = 7200
The amounts to be invested at 10% and 15% are ₹ 7,200 and ₹ 4,800.

Students can practice 10th Class Maths Solutions Telangana Chapter 4 Pair of Linear Equations in Two Variables InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables InText Questions

Think – Discuss

Question 1.
Two situations are given below. (Page No. 73)
i) The cost of 1 kg potatoes and 2 kg tomatoes was ₹ 30 on a certain day. After two days, the cost of 2 kg potatoes and 4 kg tomatoes was found to be ₹ 66.
ii) The coach of a cricket team of M.K. Nagar High School buys 3 bats and 6 balls for ₹ 3900. Later he buys one more bat and 2 balls for ₹ 1300.
Identify the unknowns in each situation. We observe that there are two unknowns in each case.
Solution:
i) Let the cost of 1 kg potatoes be ₹ x.
The cost of 1 kg tomatoes be ₹ y.
Given the cost of 1 kg potatoes and 2 kg tomatoes = ₹ 30
∴ x + 2y = 30
Given the cost of 2 kg potatoes and 4 kg tomatoes = ₹ 66
∴ 2x + 4y = 66
The pair of linear equations in two variables are x + 2y = 30 and 2x + 4y = 66.
We observe that there are two unknowns in this case.

ii) Let the cost of one bat be ₹ x.
The cost of one ball be ₹ y.
Given the cost of 3 bats and 6 balls = ₹ 3,900
3x + 6y = 3,900
Given 4 bats and 2 balls cost = ₹ 1,300
∴ 4x + 2y = 1,300
The pair of linear equations in two variables are 3x + 6y = 3,900 and 4x + 2y = 1300
We observe that there are two unknowns in this case.

Question 2.
Is a dependent pair of linear equations always consistent ? Why or why not ? (Page No. 79)
Answer:
If the lines intersect at a point gives the unique solution of the equations.
If the lines coincide then there are infinitely many solutions each point on the line being a solution. No, a dependent pair of linear equations are always consistent.

Try This

Mark the correct option in the following questions :

Question 1.
Which of the following equations is not a linear equation ?
a) 5 + 4x = y + 3
b) x + 2y = y – x
c) 3 – x = y² + 4
d) x + y = 0 (Page No. 75)
Solution:
a) 5 + 4x = y + 3
4x + 5 = y + 3
4x + 5 – y – 3 = 0
4x – y + 2 = 0
This equation is in the form of ax + by + c = 0 where a, b, c are real numbers.
So, the given equation is a linear equation.

b) x + 2y = y – x
x + 2y – y + x = 0
2x + y = 0
This equation is in the form of ax + by + c = 0 where a, b, c are real numbers.
So, the given equation is a linear equation.

c) 3 – x = y² + 4
3 – x – y² – 4 = 0
– x – y² – 1 = 0
x + y² + 1 = 0
This equation is not in the form of ax + by + c = 0 where a, b, c are real numbers.
∴ The given equation is not a linear equation.

d) x + y = 0
This equation is in the form of ax + by + c = 0 where a, b, c are real numbers.
So, the given equation is a linear equation.

Question 2.
Which of the following is a linear equation in one variable ? ( b )
a) 2x + 1 = y – 3
b) 2t – 1 = 2t + 5
c) 2x – 1 = x²
d) x² – x + 1 = 0 (Page No. 76)
Solution:
a) 2x + 1 = y – 3
⇒ 2x + 1 – y + 3 = 0
⇒ 2x – y + 4 = 0
Is a linear equation in two variables.
They are x and y.

b) 2t – 1 = 2t + 5
⇒ 2t – 1 – 2t – 5 = 0
⇒ -6 = 0 (False)
Is not a linear equation in one variable.

c) 2x – 1 = x² ⇒ x² – 2x + 1 = 0
Is a linear equation in one variable i.e., ‘x’.

d) x² – x + 1 = 0
Is a linear equation in one variable.
i.e., ‘x’.

Question 3.
Which of the following numbers is a solution for the equation
2(x + 3) = 18 ? ( b )
a) 5
b) 6
c) 13
d) 21 (Page No. 76)
Solution:
Given equation 2(x + 3) = 18
a) At x = 5; 2 (5 + 3) = 18
2 × 8 = 18
16 = 18 (False) Not a solution.

b) At x = 6; 2(6 + 3) = 18
2 × 9 = 18
18 = 18 (True)
x = 6 is a solution for the given equation.

c) At x = 13; 2(13 + 3) = 18
2 × 16 = 18
32 = 18 (False) Not a solution.

d) At x = 21; 2(21 + 3) = 18
2 × 24 = 18
48 = 18 (False) Not a solution.

Question 4.
The value of x which satisfies the equation 2x – (4 – x) = 5 – x is
a) 4.5
b) 3
c) 2.25
d) 0.5 ( c ) (Page No. 76)
Solution:
Given equation 2x – (4 – x) = 5 – x
a) At x = 4.5; 2 (4.5) – (4 – 4.5) = 5 – 4.5
9 – (-0.5) = (0.5)
9 + 0.5 = 0.5
9.5 = 0.5 (False)
∴ Value of x does not satisfies the equation.

b) At x = 3 ; 2(3) – (4 – 3) = 5 – 3
6 – 1 = 2
5 = 2 (False)
∴ Value of x does not satisfies the equation.

c) At x = 2.25 ; 2(2.25) – (4 – 2.25)
= 5 – 2.25
4.50-1.75 = 2.75
2.75 = 2.75 (True)
∴ Value of x satisfies the equation.

d) At x = 0.5 ; 2(0.5) – (4 – 0.5) = 5 – 0.5
1 – 3.5 = 4.5
– 2.5 = 4.5 (False)
∴ Value of x does not satisfies the equation.

Question 5.
The equation x – 4y = 5 has
a) no solution
b) unique solution
c) two solutions
d) infinitely many solutions (d) (Page No. 76)
Solution:
Only one equation with two unknowns (variables) we can find many solutions.

Question 6.
In the example above can you find the cost of each bat and ball ? (Page No. 79)
Solution:
No, we cannot find the cost of each bat and ball because the equations are geometrically shown by a pair of coincident lines. Every point on the line is a common solution to both the equations.

Question 7.
For what value of ‘p’ for the following pair of equations has a unique solution.
2x + py = -5 and 3x + 3y = -6 (Page No. 83)
Solution:
Given equations are 2x + py = -5 and 3x + 3y = -6
a₁ = 2 ; b₁ = p; c₁ = -5
a₂ = 3; b₂ = 3 ; c₂ = -6
\(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{2}{3}\) ; \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\) = \(\frac{p}{3}\) ; \(\frac{\mathrm{c}_1}{\mathrm{c}_2}\) = \(\frac{-5}{-6}\)
Given the pair of equations has a unique solution.
∴ \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) ≠ \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\) (∵ p ≠ 2)
If p = except 2 then the given pair of equations has a unique solution.

Question 8.
Find the value of ‘k’ for which the pair of equations 2x – ky + 3 = 0, 4x + 6y – 5 = 0 represent parallel lines. (Page No. 83)
Answer:
Given pair of equations
2x – ky + 3 = 0 and 4x + 6y – 5 = 0
a₁ = 2; b₁ = -k; c₁ = 3
a₂ = 4; b₂ = 6; c₂ = -5
Given the pair of lines are parallel.
∴ \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\)
\(\frac{a_1}{a_2}\) = \(\frac{\mathrm{b}_1}{\mathrm{~b}_2}\) ⇒ \(\frac{2}{4}\) = \(\frac{-\mathrm{k}}{6}\)
-4k = 2 × 6
-4k = 12
∴ k = \(\frac{12}{-4}\) = -3

Question 9.
For what value of ‘k’ for the pair of equations 3x + 4y + 2 = 0 and 9x + 12y + k = 0 represent coincident lines. (Page No. 83)
Solution:
Given pair of equations 3x + 4y + 2 = 0 and 9x + 12y + k = 0 (A.P.Mar.’16) (A.P. Jun.’15)
Given the pair of lines are coincident.
∴ \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\)
\(\frac{a_1}{a_2}\) = \(\frac{c_1}{c_2}\) ⇒ \(\frac{3}{9}\) = \(\frac{2}{\mathrm{k}}\) ⇒ \frac{1}{3}\(\) = \(\frac{2}{k}\)
∴ k = 3 × 2 = 6

Question 10.
For what Positive values of ‘p’ the following pair of linear equations have infinitely many solutions ? (Page No. 83)
px + 3y – (p – 3) = 0
12x + py – p = 0
Solution:
Given pair of equations are px + 3y – (p – 3) = 0 and 12x + py – p = 0
a₁ = p; b₁ = 3; c₁ = -(p – 3);
a₂ = 12; b₂ = p; c₂ = -p
Given equations has infinitely many solutions.
∴ \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{\mathrm{c}_1}{\mathrm{c}_2}\)
⇒ \(\frac{\mathrm{p}}{12}\) = \(\frac{3}{p}\) = \(\frac{-(p-3)}{-p}\)
⇒ \(\frac{\mathrm{p}}{12}\) = \(\frac{3}{p}\) = \(\frac{-\mathrm{p}+3}{-\mathrm{p}}\) ⇒ \(\frac{\mathrm{p}}{12}\) = \(\frac{3}{p}\) = \(\frac{\mathrm{p}-3}{\mathrm{p}}\)
⇒ p × p = 3 × 12
⇒ p² = 36
⇒ p = \(\sqrt{36}\) = 6

Do This

Question 1.
Solve the following systems of equations : (Page No. 79)

i) x – 2y = 0
3x + 4y = 20
Solution:
i) x – 2y = 0
-2y = -x
y = \(\frac{x}{2}\)

3x+ 4y = 20
4y = 20 – 3x
y = \(\frac{20-3 x}{4}\)

Scale :X-axis – 1 unit = 1 cm
Y – axis – 1 unit = 1 cm

The two lines are intersecting lines, meet at (4, 2).
The solution set is {(4, 2)}.

ii)
x + y = 2
2x + 2y = 4
Solution:
x + y = 2
y = 2 – x

2x + 2y = 4
2y = 4 – 2x
y = \(\frac{4-2 x}{2}\)

Scale : X – axis – 1 unit = 1 cm
Y – axis – 1 unit = 1 cm

Scale : X – axis – 1 unit = 1 cm
Y – axis – 1 unit = 1 cm
These two lines are coincident lines.
∴ There are infinitely many solutions.

iii) 2x – y = 4; 4x – 2y = 6
Solution:
2x – y = 4 ⇒ y = 2x – 4

4x – 2y = 6 ⇒ 2y = 4x – 6
⇒ y = 2x – 3

These two are parallel lines.
∴ The pair of linear equations has no solution

Scale: X – axis – 1 unit = 1 cm
Y – axis – 1 unit = 1 cm

Question 2.
Two rails on a railway track are represented by the equations.
x + 2y – 4 = 0 and 2x + 4y – 12 = 0.
Represent this situation graphically. (Page No. 79)
Solution:
x + 2y – 4 = 0; 2y = 4 – x
y = \(\frac{4-x}{2}\)
x + 2y – 4 = 0

2x + 4y – 12 = 0
⇒ 4y = 12 – 2x ⇒ 4y = 2(6 – x)
⇒ y = \(\frac{6-x}{2}\)

∴ These lines are parallel and hence no solution.

Question 3.
Check each of the given systems of equations to see if It has a unique solution, infinitely many solutions or no solution. Solve them graphically. (A.P. Mar.16’) (Page No. 83)

i) 2x + 3y = 1
3x – y = 7
Solution:
Let a₁x + b₁y + c₁ = 0 \(\simeq\) 2x + 3y – 1 = 0
a₂x + b₂y + c₂ = 0 \(\simeq\) 3x – y – 7 = 0
Now comparing their coefficients i.e.,
\(\frac{a_1}{a_2}\) and \(\frac{b_1}{b_2}\) ⇒ \(\frac{2}{3}\) ≠ \(\frac{3}{-1}\)
∴ The given lines are intersecting lines.


Scale: X – axis – 1 unit = 1 cm
Y – axis – 1 unit = 1 cm
2x + 3y = 1
⇒ 3y = 1 – 2x
⇒ y = \(\frac{1-2 x}{3}\)
∴ The system of equation has a unique solution (2, -1).

ii) x + 2y = 6
2x + 4y = 12
Solution:
From the given pair of equations,
\(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{1}{2}\) ; \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\) = \(\frac{2}{4}\) ; \(\frac{\mathrm{c}_1}{\mathrm{c}_2}\) = \(\frac{6}{12}\) = \(\frac{1}{2}\)
∴ \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\) = \(\frac{\mathrm{c}_1}{\mathrm{c}_2}\)
∴ The lines are dependent and have infinitely many solutions.
x + 2y = 6 ⇒ 2y = 6 – x
⇒ y = \(\frac{6-x}{2}\)

2x + 4y = 12
⇒ 4y = 12 – 2x ⇒ y = \(\frac{12-2 \mathrm{x}}{4}\)


Scale: X – axis – 1 unit = 1 cm
Y – axis – 1 unit = 1 cm

iii) 3x + 2y = 6
6x + 4y = 18
Solution:
From the given equations
\(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\) ; \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\); \(\frac{\mathrm{c}_1}{\mathrm{c}_2}\) = \(\frac{6}{18}\) = \(\frac{1}{3}\)
∴ \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\)
⇒ The lines are parallel and hence no solution.
3x + 2y = 6 ⇒ 2y = 6 – 3x
⇒ y = \(\frac{6-3 x}{2}\)

6x + 4y = 18 ⇒ 4y = 18 – 6x
⇒ y = \(\frac{18-6 \mathrm{x}}{4}\)


Scale : X – axis – 1 unit = 1 cm
Y – axis – 1 unit = 1 cm

Do Thiš

Question 1.
Solve each pair of equations by using the substitution method. (Page No. 88)
i) 3x – 5y = -1
x – y = -1
Solution:
Given: 3x – 5y = -1 —– (1)
x – y = -1 —– (2)
From Equation (2),
x – y = -1
x = y – 1
Substituting x = y – 1 in equation (1), we get
3(y – 1) – 5y = -1
⇒ 3y – 3 – 5y = -1
⇒ -2y = -1 + 3
⇒ 2y = -2
⇒ y = -1
Substituting y = – 1 in equation (1), we get
3x – 5(-1) = -1
3x + 5 = -1
3x = -6
x = -2
∴ The solution is (-2, -1)

ii) x + 2y = -1
2x – 3y = 12
Solution:
Given x + 2y = -1 —- (1)
2x – 3y = 12 —– (2)
From Equation (1), x + 2y = -1
⇒ x = -1 – 2y
Substituting x = -1 – 2y in equation (2), we get
2(-1 – 2y) – 3y = 12
– 2 – 4y – 3y = 12
-2 – 7y = 12
7y = -2 – 12
∴ y = \(\frac{-14}{7}\) = -2
Substituting y = -2 in equation (1), we get
x + 2(-2) = -1
x = -1 + 4
∴ x = 3
∴ The solution is (3, -2).

iii) 2x + 3y = 9
3x + 4y = 5
Solution:
Given : 2x + 3y = 9 —– (1)
3x + 4y = 5 —– (2)
From equation (1),
2x = 9 – 3y
⇒ x = \(\frac{9-3 y}{2}\)
Substituting x = \(\frac{9-3 y}{2}\) in equation (2), we get
3(\(\frac{9-3 y}{2}\)) + 4y = 5
⇒ \(\frac{27-9 y+2 \times 4 y}{2}\) = 5
⇒ 27 – 9y + 8y = 5 × 2
⇒ -y = 10 – 27
∴ y = 17
Substituting y = 17 in equation (1), we get
2x + 3 (+ 17) = 9
⇒ 2x = 9 – 51 ⇒ 2x = -42
⇒ x = -21
∴ The solution is (-21, 17).

iv) x + \(\frac{6}{y}\) = 6
3x – \(\frac{8}{y}\) = 5
Solution:
Given
x + \(\frac{6}{y}\) = 6 —- (1)
3x – \(\frac{8}{y}\) = 5 — (2)
From equation (1), x = 6 – \(\frac{6}{y}\)
Substituting x = 6 – \(\frac{6}{y}\) in equation (2), we get

Substituting y = 2 in equation (1), we get
x + \(\frac{6}{2}\) = 6
⇒ x + 3 = 6
∴ x = 3
∴ The solution is (3, 2).

v) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
Solution:
Given:
0.2x + 0.3y = 1.3 ⇒ 2x + 3y = 13 —- (1)
0.4x + 0.5y = 2.3 ⇒ 4x + 5y = 23 —- (2)
From equation (1),
2x = 13 – 3y = x = \(\frac{13-3 y}{2}\)
Substituting x = \(\frac{13-3 y}{2}\) in equation (2), we get
4(\(\frac{13-3 y}{2}\)) + 5y = 23
⇒ 26 – 6y + 5y = 23
⇒ -y + 26 = 23
⇒ y = 26 – 23 = 3
Substituting y = 3 in equation (1). we get
⇒ 2x + 3(3) = 13
⇒ 2x + 9 = 13
⇒ 2x = 13 – 9
⇒ 2x = 4
⇒ x = 2
∴ The solution is (2, 3).

vi) \(\sqrt{2}\)x + \(\sqrt{3}\)y = 0
\(\sqrt{3}\)x – \(\sqrt{8}\)y = 0
Solution:
Given:
\(\sqrt{2}\)x + \(\sqrt{3}\)y = 0 —–(1)
\(\sqrt{3}\)x – \(\sqrt{8}\)y = 0 —— (2)

∴ The solution is x = 0, y = 0.
Note: a₁x + b₁y + c = 0
a₂x + b₂y + c₂ = 0
then, x = 0, y = 0 is a solution.

Question 2.
Solve each of the following pairs of equations by the elimination method. (Page No. 89)
i) 8x + 5y = 9
3x + 2y = 4
Solution:
Given : 8x + 5y = 9 —- (1)
3x + 2y = 4 —- (2)

Substituting y = 5 in equation (1), we get
8x + 5 × 5 = 9
⇒ 8x = 9 – 25
x = \(\frac{-16}{8}\)
∴ The solution is (-2, 5).

ii) 2x + 3y = 8
4x + 6y = 7
Solution:
Given: 2x + 3y = 8 —- (1)
4x + 6y = 7 —- (2)

The lines are parallel.
∴ The pair of lines has no solution.

iii) 3x + 4y = 25
5x – 6y = -9
Solution:
Given: 3x + 4y = 25 — (1)
5x – 6y = -9 — (2)

Substituting y = 4 in equation (1), we get
3x + 4 × 4 = 25
3x = 25 – 16
∴ x = \(\frac{9}{3}\) = 3
∴ (3, 4) is the solution for given pair of lines.

Question 3.
In a competitive exam, 3 marks are to be awarded for every correct answer and for every wrong answer, 1 mark will be deducted. Madhu scored 40 marks in this exam. Had 4 marks been awarded for each correct answer and 2 marks deducted for each incorrect answer, Madhu would have scored 50 marks. How many questions were there in the test ? (Madhu attempted all the questions). Use the elimination method. (Page No. 91)
Solution:
The equations formed are
3x – y = 40 — (1)
4x – 2y = 50 — (2)

Substituting y = 5 in equation (1), we get
3x – 5 = 40
3x = 40 + 5
x = \(\frac{45}{3}\) = 15
Total number of questions = Number of correct questions + Number of wrong answers
= x + y
= 15 + 5 = 20

Question 4.
Mary told her daughter, Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be. Find the present age of Mary and her daughter. Solve by the substitution method.
(Page No. 92)
Solution:
The Equation formed are
x – 7y + 42 = 0 —- (1)
x – 3y – 6 = 0 — (2)
From (1), x = -42 + 7y
Substituting in equation (2), we get
x = -42 + 7y
⇒ -42 + 7y – 3y – 6 = 0
⇒ 4y – 48 = 0
⇒ y = \(\frac{48}{4}\) = 12
Substituting y = 12 in equation (2), we get
x – 3 × 12 – 6 = 0
x – 36 – 6 = 0
∴ x = 42

Students can practice TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise

Question 1.
Solve the following equations :

i) \(\frac{2 x}{a}\) + \(\frac{\mathbf{y}}{\mathbf{b}}\) = 2
\(\frac{x}{a}\) – \(\frac{y}{b}\) = 4
Solution:
Given
\(\frac{2 x}{a}\) + \(\frac{\mathbf{y}}{\mathbf{b}}\) = 2 —- (1)
\(\frac{x}{a}\) – \(\frac{y}{b}\) = 4 — (2)
Adding eq.(1) & (2) \(\frac{3 x}{a}\) = 6 ⇒ x = \(\frac{6 a}{3}\) = 2a
Substituting x = 2a in the equation (1), we get
\(\frac{2}{a}\)(2a) + \(\frac{\mathrm{y}}{\mathrm{b}}\) = 2
⇒ 4 + \(\frac{\mathrm{y}}{\mathrm{b}}\) = 2
⇒ \(\frac{\mathrm{y}}{\mathrm{b}}\) = -2 ⇒ y = -2b
∴ The solution (x, y) = (7, 13)

ii) \(\frac{x+1}{2}\) + \(\frac{y-1}{3}\) = 8
\(\frac{x-1}{3}\) + \(\frac{y+1}{2}\) = 9
Solution:

Substituting y = 13 in equation (1), we get
3x + 2(13) = 47 ⇒ 3x = 47 – 26
⇒ 3x = 21 ⇒ x = \(\frac{21}{3}\) = 7
∴ The solution (x, y) = (7, 13)

iii) \(\frac{x}{7}\) + \(\frac{y}{3}\) = 5
\(\frac{x}{2}\) – \(\frac{y}{9}\) = 6
Solution:

Substituting y = 9 in equation (1) we get
3x + 7(9) = 105
⇒ 3x = 105 – 63
⇒ 3x = 42
⇒ x = \(\frac{42}{3}\) = 14
∴ The solution (x, y) = (14, 9)

iv) \(\sqrt{3}\)x – \(\sqrt{2}\)y = \(\sqrt{3}\)
\(\sqrt{5}\)x + \(\sqrt{3}\)y = \(\sqrt{3}\)
Solution:

v) \(\frac{a x}{\mathbf{b}}\) – \(\frac{b y}{a}\) = a + b
ax – by = 2ab
Solution:
Given \(\frac{a x}{\mathbf{b}}\) – \(\frac{b y}{a}\) = a + b —- (1)
ax – by = 2ab —- (2)

Substituting y = -a in equation (2), we get
ax – b(-a) = 2ab
⇒ ax + ab = 2ab
⇒ ax = 2ab – ab
∴ x = \(\frac{a b}{a}\) = b
∴ The solution (x, y) = (b, -a)

vi) 2^x + 3^y = 17
2^x+2 – 3^y+1 = 5
Solution:
Given, 2^x + 3^y = 17 and
2^x+2 – 3^y+1 = 5
Take 2^x = a and 3^y = b then the give equations reduce to
2^x + 3^y = 17 —- (1)
2^x.2² – 3^y. 3 = 5 ⇒ 4a – 3b = 5 —– (2)

Substituting b = 9 in equation (1), we get
a + 9 = 17 ⇒ a = 17 – 9 = 8
But a = 2^x = 8 and b = 3^y = 9
⇒ 2^x = 2³
⇒ x = 3

⇒ 3^y = 3²
⇒ y = 2
∴ The solution (x, y) is (3, 2).

Question 2.
Animals in an experiment are to be kept on a strict diet. Each animal is to receive among other things 20g of protein and 6g of fat. The laboratory technicians purchased two food mixes, A and B. Mix A has 10% protein and 6% fat. Mix B has 20% protein and 2% fat. How many grams of each mix should be used?
Solution:
Let x gms of mix A and y gms of mix B are to be mixed, then


∴ y = \(\frac{300}{5}\) = 60 gm
Substituting y = 60 in equation (1), we get
x + 2 × 60 = 200 ⇒ x + 120 = 200
⇒ 200 – 120 = 80 gm
∴ Quantity of mix. A = 80 gms.
Quantity of mix. B = 60 gms.